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Find the smallest positive \(N \) such that 

\(N\equiv3 (mod 4) \)

\(N\equiv2 (mod 5) \)

\(N\equiv6 (mod 7) \)

 Aug 9, 2018
 #1
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N mod 4 = 3
N mod 5 = 2
N mod 7 = 6 

 

4A+3=5B+2=7C+6
A = 6, B = 5, C = 3
Therefore:
4 x 6 + 3 = 27

 

Since LCM of [4, 5, 7] = 140, then:
N =140D + 27, where D =0, 1, 2, 3.......etc.

 Aug 9, 2018

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